Single Variable Calculus, Chapter 4, 4.2, Section 4.2, Problem 12

Verify that the function $f(x) = x^3 + x - 1$ satisfies the hypothesis of the Mean Value Theorem on the interval $[0,2]$. Then find all the numbers $c$ that satisfy the conclusion of the Mean Value Theorem.
We know that $f(x)$ is a polynomial function that is continuous everywhere and its derivative, $f'(x) = 3x^2 + 1$ is a quadratic function that is continuous on every values of $x$ as well. Hence, $f(x)$ is continuous on the closed interval $[0,2]$ and differentiable on the open interval $(0,2)$.

Now solving for $c$, we have...

$\begin{equation} \begin{aligned} f'(c) &= \frac{f(b) - f(a)}{b - a }\\ \\ f'(c) &= \frac{\left[ (2)^3 + 2 - 1\right] - \left[ (0)^3 + 0 - 1\right]}{2- 0}\\ \\ f'(c) &= 5 \end{aligned} \end{equation}$

but $f'(x) = 3x^2 + 1$, so $f'(c) = 3(c)^2 +1$

$\begin{equation} \begin{aligned} 3c ^2 + 1 &= 5\\ \\ 3c^2 &= 4\\ \\ c^2 &= \frac{4}{3}\\ \\ c &= \pm \sqrt{\frac{4}{3}} \end{aligned} \end{equation}$

We got two values of $c$ but the function is defined only on interval $[0,2]$. Hence, $\displaystyle c = - \sqrt{\frac{4}{3}}$ is disregarded. Therefore, $\displaystyle c = \sqrt{\frac{4}{3}}$